Algebra - Surds & Indices - Previous Year CAT/MBA Questions

You can practice all previous year OMET questions from the topic Algebra - Surds & Indices. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.

1. XAT 2019 QADI | Algebra - Surds & Indices

If $\sqrt[3]{7^{a} \times (35)^{(b + 1)} \times (20)^{(c + 2)}}$ is a whole number then which one of the statements below is consistent with it?

A.
a = 3, b = 2, c = 1
B.
a = 3, b = 1, c = 1
C.
a = 2, b = 1, c = 2
D.
a = 2, b = 1, c = 1
E.
a = 1, b = 2, c = 2

2. IIFT 2019 QA | Algebra - Surds & Indices

If x = 8 - $\sqrt{32}$ and y = 2 + √2, then ${(x + \frac{1}{y})}^{2}$ is given by:

A.
16x²/25
B.
64x²/81
C.
25y²/16
D.
81x²/64

Answer: Option D

Explanation :

x = 8 − √32 = 8 − 4√2 = 4(2 − √2)

$\frac{1}{y}$ = $\frac{1}{2 + \sqrt{2}}$

On rationalising, we get

$\frac{1}{y}$ = $\frac{1}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}}$ = $\frac{2 - \sqrt{2}}{2}$ = $\frac{x}{8}$

∴ x + 1/y = x + x/8 = 9x/8

⇒ ${(x + \frac{1}{y})}^{2}$ = ${(x + \frac{x}{8})}^{2}$ = ${(\frac{9 x}{8})}^{2}$ = $\frac{81 x^{2}}{64}$

Hence option (d).

3. IIFT 2016 QA | Algebra - Surds & Indices

The highest number amongst $\sqrt{2}, \sqrt[3]{3},$ and $\sqrt[4]{4}$

A.
$\sqrt{2}$
B.
$\sqrt[3]{3}$
C.
$\sqrt[4]{4}$
D.
None of the above

4. IIFT 2015 QA | Algebra - Surds & Indices

The value of x for which the equation $\sqrt{4 x - 9} + \sqrt{4 x + 9} = 5 + \sqrt{7}$ will be satisfied is:

A.
1
B.
2
C.
3
D.
4

5. IIFT 2015 QA | Algebra - Surds & Indices

The simplest value of the expression ${\{\frac{4^{p + \frac{1}{4}} \times \sqrt{2 \times 2^{p}}}{2 \times \sqrt{2^{- p}}}\}}^{\frac{1}{p}}$ is:

A.
4
B.
8
C.
4^p
D.
8^p

Answer: Option B

Explanation :

$\{\frac{4^{p + \frac{1}{4}} \times \sqrt{2 \times 2^{p}}}{2 \times \sqrt{2^{- p}}}\} = \{\frac{4^{p} \times 4^{\frac{1}{4}} \times \sqrt{2} \times \sqrt{2^{p}}}{2 \times \sqrt{2^{- p}}}\} = \{\frac{2^{2 p} \times \sqrt{2} \times \sqrt{2} \times \sqrt{2^{p}}}{2 \times \sqrt{2^{- p}}}\} = 2^{2 p} \times \sqrt{2^{p}} \times \sqrt{2^{p}} = 2^{3 p}$

${\{\frac{4^{p + \frac{1}{4}} \times \sqrt{2 \times 2^{p}}}{2 \times \sqrt{2^{- p}}}\}}^{\frac{1}{p}} = {(2^{3 p})}^{\frac{1}{p}} = 8$

Hence, option 2.

Algebra - Surds & Indices - Previous Year CAT/MBA Questions

Feedback